TITLE: What's the smallest exponent to give the identity in $S_n$?
QUESTION [6 upvotes]: Let $S_n$ denote the symmetric group on $n$ letters. We know that $\tau^{n!} = e$ for any element $\tau \in S_n,$ where $e$ denotes the identity element. Can we find a smaller positive integer $m$ with this property? That is, can we find a positive integer $m < n!$ such that $$\tau^m = e$$ for all $\tau \in S_n$?

REPLY [6 votes]: The answer is yes for $n \ge 4$. When you write a permutation $\tau$ as a product of disjoint cycles, its period $p$ is the lcm of the lengths of the cycles. So if $2^k$ divides $p$, $\tau$ contains a $2^k$ cycle and so $2^k \le n$. For $n \ge 4$, $n! = 2^rs$ where $s$ is odd and $2^r > n$, so $p$ will divide $m = 2^ks$, where $k$ is the largest integer such that $2^k \le n$. Similarly you can reduce the exponents of other prime divisors of $n!$ as soon as $n$ is large enough